can someone please help me on these questions??

Question

1. log_5 x+3=log_5 (x-20)+4

2. log_2 x+5=8-log_2 (x+7)

3. log_x 9=-2

Answer 1

I am not certain what you mean but I will assume that log_5 means the log to the base 5, etc.

1. log_5(x + 3) = log_5(x - 20) + 4

log_5(x + 3) = log(x + 3)/log(5)
log_5(x - 20) = log(x - 20)/log(5)

log(x + 3) = log(x - 20) + 4log(5)
log(x + 3) - log(x - 20) = log[(x + 3)/(x - 20)] = 4log(5)
(x + 3)/(x - 20) = 10^[4log(5)] = 5^4 = 625
x(1 - 625) = -20(625) - 3
x = 12503/624
x = 20.03686

Check:
log_5(x + 3) =1.9492
log_5(x - 20) + 4 =1.9492

2.log_2(x + 5) = 8 - log_2(x + 7)

log_2(x + 5) = log(x + 5)/log(2)
log_2(x + 7) = log(x + 7)/log(2)

log(x+5) = 8log(2) - log(x + 7)
log[(x + 5)(x + 7)] = 8log(2) = log(2^8)
(x + 5)(x + 7) = 2^8 = 256
x^2 + 12x + 35 = 256
x^2 + 12x + 36= 257
(x + 6) = SQRT(257)
x = SQRT(257) - 6
x = 10.03122

Check:
log_2(x + 5) = 3.91
8 - log_2(x + 7) = 3.91

3. log_x(9) = -2
log(9)/log(x) = -2
log(x) = (-1/2)log(9) = log[9^(-1/2)]
x = 9^(-1/2) = 1/3

Check:
log_1/3(9) = -2

Answer 2

formula used
(i) log_a b =c then a^c =b
(ii) log_a b + log_a c = log_a (b*c)
(iii) log_a b - log_a c = log_a (b/c)

1)log_5 x+3=log_5 (x-20)+4
log_5 x - log_5 (x-20) = 4-3
log_5 (x / (x-20) ) = 1
x / (x-20) = 5^1
x / (x-20) = 5
x = 5(x-20)
x = 5x - 100
x - 5x = -100
-4x = -100
x = -100 / -4 =25
2)log_2 x+5=8-log_2 (x+7)
log_2 x + log_2 (x+7) = 8-5
log_2 (x * (x+7)) =3
log_2 (x^2 + 7x) =3
x^2 + 7x = 2^3
x^2 + 7x =8
x^2 + 7x -8 =0
(x-1)(x+8) =0
x= 1,-8 as negative value is not accepted for log so
x =1

3) log_x 9=-2
so, x^-2 = 9
x = 9^(-1/2) = (3^2)^(-1/2) = 3^(2* -1/2) = 3^-1 = 1/3